9702 w10 ms_all

UNIVERSITY OF CAMBRIDGE INTERNATIONAL EXAMINATIONS

GCE Advanced Subsidiary Level and GCE Advanced Level

MARK SCHEME for the October/November 2010 question paper

for the guidance of teachers

9702 PHYSICS

9702/11 Paper 1 (Multiple Choice), maximum raw mark 40

Mark schemes must be read in conjunction with the question papers and the report on the examination.

• CIE will not enter into discussions or correspondence in connection with these mark schemes. CIE is publishing the mark schemes for the October/November 2010 question papers for most IGCSE, GCE Advanced Level and Advanced Subsidiary Level syllabuses and some Ordinary Level syllabuses.

Page 2 Mark Scheme: Teachers’ version Syllabus Paper

GCE A/AS LEVEL – October/November 2010 9702 11

© UCLES 2010

Question Number

Key Question Number

Key

1 D 21 A

2 C 22 B

3 B 23 A

4 A 24 C

5 B 25 A

6 C 26 A

7 B 27 D

8 A 28 A

9 C 29 D

10 B 30 B

11 B 31 C

12 B 32 A

13 A 33 B

14 B 34 C

15 D 35 C

16 B 36 B

17 D 37 A

18 D 38 C

19 D 39 D

20 D 40 C





9702 PHYSICS






© UCLES 2010

Question Number

Key Question Number

Key

1 A 21 A

2 B 22 A

3 A 23 C

4 C 24 C

5 D 25 D

6 D 26 C

7 D 27 B

8 A 28 A

9 C 29 A

10 B 30 D

11 B 31 A

12 C 32 C

13 D 33 D

14 D 34 B

15 D 35 B

16 B 36 B

17 C 37 B

18 C 38 D

19 C 39 C

20 D 40 B





9702 PHYSICS






© UCLES 2010

Question Number

Key Question Number

Key

1 B 21 B

2 B 22 D

3 C 23 A

4 D 24 A

5 A 25 C

6 A 26 D

7 C 27 A

8 B 28 D

9 B 29 B

10 B 30 A

11 B 31 C

12 C 32 C

13 D 33 C

14 A 34 B

15 B 35 A

16 D 36 A

17 D 37 B

18 B 38 C

19 A 39 C

20 D 40 D





9702 PHYSICS

9702/21 Paper 2 (AS Structured Questions), maximum raw mark 60

This mark scheme is published as an aid to teachers and candidates, to indicate the requirements of the examination. It shows the basis on which Examiners were instructed to award marks. It does not indicate the details of the discussions that took place at an Examiners’ meeting before marking began, which would have considered the acceptability of alternative answers.




GCE AS/A LEVEL – October/November 2010 9702 21

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1 (a) length, current, temperature, amount of substance, (luminous intensity) any three, 1 each B3 [3] (b) (i) F: kg m s–2 B1

ρ: kg m–3 B1 v: m s–1 B1 [3] (ii) some working e.g. kg m s–2 = m2 kg m–3 (m s–1)k M1 hence k = 2 A1 [2] 2 (a) (i) horizontal speed constant at 8.2 m s–1 C1 vertical component of speed = 8.2 tan 60° M1 = 14.2 m s–1 A0 [2] (ii) 14.22 = 2 × 9.8 × h (using g = 10 then –1) C1 vertical distance = 10.3 m A1 [2] (iii) time of descent = 14.2 / 9.8 = 1.45 s C1 x = 1.45 × 8.2 = 11.9 m A1 [2] (b) (i) smooth path curved and above given path M1 hits ground at more acute angle A1 [2] (ii) smooth path curved and below given path M1 hits ground at steeper angle A1 [2] 3 (a) force = rate of change of momentum (allow symbols if defined) B1 [1]

(b) (i) ∆ρ = 140 × 10–3 × (5.5 + 4.0) C1 = 1.33 kg m s–1 A1 [2] (ii) force = 1.33 / 0.04 M1 = 33.3 N A0 [1] (c) (i) taking moments about B C1 (33 × 75) + (0.45 × g × 25) = FA × 20 C1 FA = 129 N A1 [3] (ii) FB = 33 + 129 + 0.45g C1 = 166 N A1 [2]



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4 (a) (i) F / A B1 [1]

(ii) ∆L / L B1 [1]

(iii) allow FL / A∆L B1 [1]

(iv) allow ρL / A or ρ (L + ∆L) / A B1 [1]

(b) (i) ∆L = FL / EA = (30 × 2.6) / (7.0 × 1010 × 3.8 × 10–7) M1 = 2.93 × 10–3 m = 2.93 mm A0 [1]

(ii) ∆R = ρ∆L / A C1 = (2.6 × 10–8 × 2.93 × 10–3) / (3.8 × 10–7)

= 2.0 × 10–4 Ω A1 [2] (c) change in resistance is (very) small M1 so method is not appropriate A1 [2] 5 (a) when a wave passes through a slit / by an edge M1 the wave spreads out / changes direction A1 [2] (b) diagram: wavelength unchanged M1 wavefront flat at centre, curving into geometrical shadow A1 [2]

(c) d sin θ = nλ C1

for θ = 90° 1 / (650 × 103) = n × 590 × 10–9 M1 n = 2.6 number of orders is 2 A1 [3] (d) intensity / brightness decreases (as order increases) B1 [1]

6 (a) (i) either P = V 2 / R or P = VI and V = IR C1

R = 4.0 Ω A1 [2] (ii) sketch vertical axis labelled appropriately B1 (straight) line from origin then curved in correct direction B1 line passes through 12 V, 3.0 A B1 [3] (b) (i) 2.0 kW A1 [1] (ii) 0.5 kW A1 [1] (iii) total resistance = 3R / 2 C1 power = 0.67 kW A1 [2]



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7 (a) either different forms of same element or nuclei have same number of protons M1 different numbers of neutrons (in the nucleus) A1 [2] (b) (i) proton number conserved B1 nucleon number conserved B1 mass-energy conserved B1 [3] (ii) 1. Z = 36 A1 [1] 2. x = 3 A1 [1]





9702 PHYSICS






GCE A LEVEL – October/November 2010 9702 22

© UCLES 2010

1 (a) (i) scalar quantity has magnitude (allow size) B1 vector quantity has magnitude and direction B1 [2] (ii) 1. temperature: scalar B1 [1] 2. acceleration: vector B1 [1] 3. resistance: scalar B1 [1] (b) either triangle / parallelogram with correct shape C1 tension = 14 .3 N (allow ± 0.5 N) A2 [3] (if > ±0.5 N but ≤ ±1 N, allow 1 mark) or R = 25 cos 35° (C1) T = R tan 35° (C1) T = 14.3 N (A1) or T = 25 sin 35° (C2) T = 14.3 N (A1) or R and T resolved vertically and horizontally (C2) leading to T = 14.3 N (A1) 2 (a) (i) VH = 12.4 cos 36° (= 10.0 m s–1) C1 distance = 10.0 × 0.17 = 1.7 m A1 [2] (ii) VV = 12.4 sin 36° (= 7.29 m s–1) C1 h = 7.29 × 0.17 – ½ × 9.81 × 0.172 C1 = 1.1 m A1 [3] (b) smooth curve with ball hitting wall below original B1 smooth curve showing rebound to ground with correct reflection at wall B1 [2] 3 (a) point at which (whole) weight (of body) (allow mass for weight) M1 appears / seems to act ... (for mass need ‘appears to be concentrated’) A1 [2] (b) (i) point C shown at centre of rectangle ± 5 mm B1 [1] (ii) arrow vertically downwards, from C with arrow starting from the same margin of error as in (b)(i) B1 [1] (c) (i) reaction / upwards / supporting / normal reaction force M1 friction M1 force(s) at the rod A1 [3] (ii) comes to rest with (line of action of) weight acting through rod allow C vertically below the rod B1 so that weight does not have a moment about the pivot / rod B1 [2]



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4 (a) energy = average force × extension B1 = ½ × F × x B1 (Hooke’s law) extension proportional to (applied) force B1 hence F = kx B1 so E = ½kx2 A0 [4] (b) (i) correct area shaded B1 [1] (ii) 1.0 cm2 represents 1.0 mJ or correct units used in calculation C1 ES = 6.4 ± 0.2 mJ A2 [3] (for answer > ±0.2 mJ but ≤ ±0.4 mJ, then allow 2/3 marks) (iii) arrangement of atoms / molecules is changed B1 [1] 5 (a) (i) distance (of point on wave) from rest / equilibrium position B1 [1] (ii) distance moved by wave energy / wavefront during one cycle of the source or minimum distance between two points with the same phase or between adjacent crests or troughs B1 [1] (b) (i) T = 0.60 s B1 [1]

(ii) λ = 4.0 cm B1 [1]

(iii) either v = λ/T or v = fλ and f = 1/T C1 v = 6.7 cm s–1 A1 [2] (c) (i) amplitude is decreasing M1 so, it is losing power A1 [2] (ii) intensity ~ (amplitude)2 C1 ratio = 2.02 / 1.12 C1 = 3.3 A1 [3]

6 (a) (i) at 22.5 °C, RT = 1600 Ω or 1.6 kΩ C1

total resistance = 800 Ω A1 [2] (ii) either use of potential divider formula or current = 9 / 2000 (4.5 mA) C1 V = (0.8/2.0) × 9 V = (9/2000) × 800 = 3.6 V = 3.6 V A1 [2] (b) (i) total resistance = 4/5 × 1200 C1

= 960 Ω A1 [2] (ii) for parallel combination, 1/960 = 1/1600 + 1/RT

RT = 2400 Ω / 2.4 kΩ C1

temperature = 11 °C A1 [2]



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(c) e.g. only small part of scale used / small sensitivity B1 non-linear B1 [2] (any two sensible suggestions, 1 each, max 2)

7 (a) (i) most α-particles were deviated through small angles B2 [2] (allow 1 mark for ‘straight through’ / undeviated)

(ii) small fraction of α-particles deviated through large angles M1

greater than 90° (allow rebound back) A1 [2]

(b) e.g. β-particles have a range of energies

β-particles deviated by (orbital) electrons

β-particle has (very) small mass (any two sensible suggestions, 1 each, max 2) B2 [2]

Do not allow β-particles have negative charge or β-particles have high speed





9702 PHYSICS







© UCLES 2010

1 (a) allow 0.05 mm → 0.15 mm B1 [1] (b) allow 0.25 s → 0.5 s B1 [1] (c) allow 8 N → 12 N B1 [1] ignore number of significant figures 2 crystalline: atoms / ions / particles in a regular arrangement / lattice long range order / orderly pattern B1 (lattice) repeats itself (1) polymer: long chain molecules / chains of monomers B1 some cross-linking between chains / tangled chains (1) amorphous: disordered arrangement of molecules / atoms / particles B1 any ordering is short-range (1) (three ‘B’ marks plus any other 2 marks) B2 [5] 3 connect microphone / (terminals of) loudspeaker to Y-plates of c.r.o. B1 adjust c.r.o. to produce steady wave of 1 (or 2) cycles / wavelengths on screen B1

measure length of cycle / wavelength λ and note time-base b M1

frequency = 1 / λb A1 [4] (assume b is measured as s cm–1, unless otherwise stated) (if statement is ‘measure T, f = 1/T’ then last two marks are lost) 4 (a) acceptable straight line drawn (touching every point) B1 [1] (b) the distance fallen is not d C1 d is the distance fallen plus the diameter of the ball A1 [2] (‘d is not measured to the bottom of the ball’ scores 2/2) (c) (i) diameter: allow 1.5 ± 0.5 cm (accept one SF) A1 [1] no ecf from (a) (ii) gradient = 4.76, ± 0.1 with evidence that origin has not been used C1 gradient = g / 2 C1 g = 9.5 m s

–2 A1 [3]



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5 (a) (i) Fig. 5.2 B1 [1] (ii) Fig. 5.3 B1 [1] (b) kinetic energy increases from zero then decreases to zero B1 [1]

(c) (i) ∆EP = mg∆h / mgh C1 = 94 × 10–3 × 9.8 × 2.6 × 10–2 using g = 10 then –1 = 0.024 J A1 [2] (ii) either 0.024 = ½ k × (2.6 × 10–2)2 or ½ kd2 = ½k × (2.6 × 10–2)2 – ½kd2 C1 0.012 = ½k × d2 kd2 = ½k × (2.6 × 10–2)2 C1 d = 0.018 m d = 0.018 m = 1.8 cm = 1.8 cm A1 [3] 6 (a) when two (or more) waves meet (at a point) B1 (resultant) displacement is (vector) sum of individual displacements B1 [2]

(b) (i) λ = ax / D (if no formula given and substitution is incorrect then 0/3) C1 590 × 10–9 = (1.4 × 10–3 × x) / 2.6 C1 x = 1.1 mm A1 [3]

(ii) 1. 180° (allow π if rad stated) A1 [1] 2. at maximum, amplitude is 3.4 units and at minimum, 0.6 units C1

intensity ~ amplitude2 allow I ~ a2 C1 ratio = 3.42 / 0.62 = 32 A1 [3] 7 (a) (i) path: reasonable curve upwards between plates B1 straight and at a tangent to the curve beyond the plates B1 [2] (ii) 1. (F =) E.g B1 [1] 2. (t =) L / v B1 [1] (b) (i) total momentum of a system remains constant or total momentum of a system before a collision equals total momentum after collision M1 provided no external force acts on the system A1 [2] (do not accept ‘conserved’ but otherwise correct statement gets 1/2)

(ii) (∆p =) EqL / v allow ecf from (a)(ii) B1 [1] (iii) either charged particle is not an isolated system M1 so law does not apply A1 [2] or system is particle and ‘plates’ (M1)

equal and opposite ∆p on plates / so law applies (A1)



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8 (a) (i) either P = V2 / R or I = 1200 / 230 or 5.22 C1

R = (230 × 230) / 1200 R = 2302 / 1200 or R = 230 / 5.22 M1

= 44.1 Ω = 44.1 Ω A0 [2]

(ii) R = ρL / A C1

= (1.7 × 10–8 × 9.2 × 2) / (π × 0.45 × 10–32) M1

= 0.492 Ω A0 [2] (b) current = 230 /44.6 C1 power = (230 /44.6)2 × 44.1 C1 = 1170 W A1 [3] (allow full credit for solution based on potential divider) (c) e.g. less power dissipated in the heater / smaller p.d. across heater / more power loss in cable / current lower B1 cable becomes heated / melts B1 [2] (any two sensible suggestions, 1 each, max 2)

9 (a) nucleus emits α-particles or β-particles and/or γ-radiation B1 to form a different / more stable nucleus B1 [2] (b) (i) fluctuations in count rate (not ‘count rate is not constant’) B1 [1] (ii) no effect B1 [1]

(iii) if the source is an α-emitter B1

either α-particles stopped within source (and gain electrons)

or α-particles are helium nuclei B1 [2] allow 1/2 for ‘parent nucleus gives off radiation to form daughter nucleus’





9702 PHYSICS

9702/31 Paper 31 (Advanced Practical Skills 1), maximum raw mark 40






© UCLES 2010

1 (a) (i) No help from Supervisor. [1] (ii) Values of a and b with consistent units to the nearest mm. [1] (b) Six sets of readings of a, b and R scores 5 marks, five sets scores 4 marks etc. [5]

Incorrect trend then –1. Correct trend b/a increases, R increases. Major help from supervisor –1.

Range: used R = 8000 Ω or 7000 Ω. [1]

Column headings (R/Ω, a/m, b/m, b/a). [1] Must have R and either b/a or a and b columns. Each column heading must contain a quantity and a unit where appropriate. Ignore any units in the body of the table. There must be some distinguishing mark between the quantity and the unit (solidus is

expected but accept, for example, R (Ω). Consistency of presentation of readings. [1] All values of raw a and b must be given to the nearest mm. Significant figures. [1] Significant figures for b/a must be the same as, or one more than, the least number of s.f. used in a or b. Correct calculation of b/a. [1]

(c) (i) Axes: [1]

Sensible scales must be used. No awkward scales (e.g. 3:10). Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity which is being plotted. Ignore units. Scale markings should be no more than three large squares apart. All observations must be plotted. Ignore any plot off the grid. [1] Write a ringed total of plotted points. Ring and check a suspect point. Work to an accuracy of half a small square. Do not accept blobs (points with diameter > 0.5 small square).

(ii) Line of best fit. [1]

Judge by balance of at least 5 trend points about candidate’s line. There must be an even distribution of points either side of the line along the full length. Line must not be kinked. Do not allow lines thicker than half a small square. Quality. [1]

Scatter of points must be less than ± 200 Ω in the R – axis about a straight line. All points in the table must be plotted (at least 5) for this mark to be awarded.

(iii) Gradient. [1]

The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square.



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(d) Gradient = X

1 [1]

Value of X in range 3000–3600 Ω with unit.

(e) a

b = 1 [1]

Correct reading off graph. [1] [Total: 20] 2 (c) (ii) Measurement of h to nearest mm with consistent unit. 0.900 m < h < 1.100 m [1] (d) (ii) Value of mA – mB = 20 g with consistent unit. [1] (iii) Value of t with unit. t < 5 seconds [1]

Evidence of repeated measurements of t. [1]

(e) Absolute uncertainty in t in range 0.1–0.6 s. [1]

If repeated readings have been taken, then the uncertainty can be half the range. Correct method of calculation to get percentage uncertainty. [1]

(f) Second value of mA – mB = 40 g [1]

Second value of t. [1] Quality: second value of t < first value of t. [1]

(g) (i) Values of k calculated correctly. [1] (ii) Justification of sf in k linked t and (mA – mB) or mA and mB or masses. [1] (iii) Valid conclusion based on the calculated values of k. [1]

Candidate must test against a stated criterion.



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(h) Identifying limitations marks and suggesting improvements

(i) Limitations [4] (ii) Improvements [4] Do not credit

Ap Two readings are not enough (to draw a conclusion)

As Take more readings and plot a graph/calculate more values of k.

One reading/few readings/take more readings and average.

Bp Masses hit each other/ masses slipping off.

Bs Use larger pulley/method of securing masses to hanger.

Cp Uncertain starting position Cs Method of fixing rule e.g. clamp rule/electromagnetic release mechanism

Dp Difficult to measure time as time short/reaction time large compared with time.

Ds Drop through greater height/ expand on trap door mechanism/ light gate with timer/motion sensor with data logger/video timer with timer.

Ep Friction at pulley Es Lubricate pulley Friction between pulley and string

Fp Retort stand moves Fs Method of fixing to the bench e.g. clamp/add weights

Gp Mass (values) not accurate

Gs Use balance/method of measuring mass

Do not credit parallax error. [Total: 20]





9702 PHYSICS







© UCLES 2010

1 (a) (ii) Value of raw h to the nearest mm (unit needed). h > 20 cm. [1] (b) Evidence of repeat times: of one swing repeated several times or the time for a number

of swings recorded at least once (not fixed time and count n). [1] Value of 0.5 < T < 3 s.

(c) Six sets of readings of x and T scores 5 marks, five sets scores 4 marks etc. [5]

Incorrect or no trend then –1 (Correct trend x increases, T2 decreases). SH –1. Write a ringed total next to the table. Maximum value of x at least h/2. [1] Column headings (x / m, x / mm, T / s, T2/s2). [1] Must have x and T2 columns. Each column heading must contain a quantity and a unit. Ignore any units in the body of the table. There must be some distinguishing mark between the quantity and the unit (solidus is expected but accept, for example, x (m)). Consistency of presentation of raw readings. [1] All values of raw x must be given to the nearest mm and all values of raw time to the same number of d.p. (either 1 or 2). Significant figures. [1] Significant figures for T2 must be the same as, or one more than, the least number of significant figures used in the raw time data. Also if raw time is given to the nearest hundredth of a second accept one less significant figure in T2. Correct calculation of T2. Do not allow t2. [1]



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(d) (i) Axes: [1] Sensible scales must be used. No awkward scales (e.g. 3:10). Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity which is being plotted. Ignore units. Scale markings should be no more than three large squares apart. All observations must be plotted on the grid. [1] Write a ringed total of plotted points. Ring and check a suspect plot. Work to an accuracy of half a small square. Do not accept blobs (points with diameter > 0.5 small square).

(ii) Line of best fit. [1]

Judge by balance of at least 5 points about the candidate’s line. There must be an even distribution of points either side of the line along the full length. Line must not be kinked. Do not allow lines thicker than half a small square. Quality. [1] Scatter of points must be less than ± 1 cm (to scale) in the x (cm) direction of a straight line. All points in table must be plotted (at least 5) for this mark to be awarded.

(iii) Gradient. [1]

Negative sign must be seen on answer line consistent with graph. The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square. Intercept. [1] Either: Check correct read-off from a point on the line and substitution into y = mx + c. Read off must be accurate to half a small square. Allow ecf of gradient value. Or: Check read-off of intercept directly from the graph.

(e) Value of gradient

intercept

−

−

=

y

B

A (Expect value to be approximately equal to h). [1]

Unit for A/B correct (e.g. m) consistent with value. [1] Allow candidate’s value 0.5 h < A/B < 1.5 h.

[Total: 20]



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2 Measurement of dA in range 0.20 mm < dA < 0.40 mm to nearest 0.01 mm or 0.001 mm with consistent unit. If OOR allow SV ± 0.10 mm. [1] Evidence of repeated measurements of d (or in (e)). [1]

(c) (i) Measurement of L to nearest mm with consistent unit. [1] (ii) Absolute uncertainty in L is 2 mm–10 mm. [1]

If repeated readings have been taken, then the uncertainty can be half the range. Correct method of calculation to get percentage uncertainty. [1]

(d) (ii) Measurement of VA. Any supervisor’s help –1. [1] (e) Value of dB. Major help from supervisor –1. [1] (f) (ii) Measurement of VB to at least nearest 0.1 V with unit. V < 2 V. If > 2 V check SV. [1]

Quality: VB < VA. [1]

(g) (i) Values of k calculated correctly. [1] (ii) Justification of sf in k linked to L and d and V. [1] (iii) Valid conclusion based on the calculated values of k. [1]




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(h)


Ap Two readings are not enough (to draw a conclusion.

As Take more readings and plot a graph/calculate more values of k.

One reading/ few readings/ take more readings and average.

Bp Difficult to measure length because (give a reason) e.g. clips have a width/ clip slips. Difficult to make L the same (for both experiments).

Bs Use sliding jockeys/narrower clips/ solder contacts/use longer wire (to reduce % error).

Cp Voltmeter scale not sensitive enough/not precise enough/only reads to 0.1 or 0.05 V.

Cs Use digital voltmeter/use a voltmeter that reads to 0.01 V.

Voltmeter not accurate enough. More accurate voltmeter.

Dp Wires kinked/Wires not straight/Difficult to keep wire straight/difficult to prevent short circuiting.

Ds Method of keeping wire (during experiment) straight e.g. tape to ruler, hang weights off end, clamp wire.

Parallax error.

Ep Difficult to make I the same (for both experiments).

Es Method to obtain continuous variation in the current e.g. (slide wire) potentiometer/potential divider/finer wire rheostat/longer rheostat.

Fp Contact resistance/ fluctuating ammeter or voltmeter readings.

Fs Method of cleaning contacts e.g. sand clips. Tighten clips.

Ignore reference to parallax error, zero error on meters, heating effects of wire, cell runs

down, video the experiment. [Total: 20]





9702 PHYSICS







© UCLES 2010

1 (c) Measurements for h1 and h2 to nearest mm [1] Check raw values if readings are repeated. The difference between h1 and h2 is < 2 mm. [1] (d) (iii) Six sets of readings of n, h1 and h2 scores 5 marks, five sets scores [5] 4 marks etc. Incorrect trend then –1. Help from supervisor then –1. Range – [1] n values must include 10 or greater. Column headings – [1] Each column heading must contain a quantity and a unit where appropriate. There must be some distinguishing mark between the quantity and the unit. E.g. h1/cm or h1(cm) but not 1/((h1 – h1)/cm). Consistency of presentation of raw readings – [1] All values of h1 and h2 must be given to the same precision. Significant figures – [1] S.f. for 1/(h1 – h2) must be the same as, or one more than, the s.f. in the difference

(h1 – h2). Calculation – [1] 1/(h1 – h2) calculated correctly. (Graph) Axes – [1] Sensible scales must be used, no awkward scales (e.g. 3:10). Scales must be

chosen so that the plotted points must occupy at least half the graph grid in both x and y directions.

Scales must be labelled with the quantity which is being plotted. Ignore units. Scale markings must be no more than 3 large squares apart. Plotting of points – [1] All observations must be plotted. Do not accept blobs (points with diameter > half a small square). Ring and check a suspect plot. Tick if correct. Re-plot if incorrect. Work to an accuracy of half a small square. Line of best fit – [1] Judge by balance of at least 5 trend points about the candidate's line. There must

be an even distribution of points either side of the line along the full length. Line must not be kinked. Quality – [1] Scatter of points must be less than ±0.02 on the 1/n axis about the examiner’s line. All points must be plotted (at least 5) for this mark to be scored.



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(e) (iii) Gradient The hypotenuse must be at least half the length of the drawn line. [1] Both read-offs must be accurate to half a small square. Intercept [1] Check that the read-off or the method of calculation is correct. (f) Value of a = value of gradient and value of b = value of intercept. [1] Do not allow a value presented as a fraction. Units for a and b are correct. [1] E.g. cm–1 or m–1 but must be consistent with the values. Allow no unit for b if b = 0. [Total: 20] 2 (a) (i) Value of d in range 5 cm to 15 cm. [1] Help from supervisor then –1. Evidence of repeated measurements of d. [1] (ii) Correct calculation of A. [1]

Do not allow a value in terms of π. (b) (i) Measurement for x in range 0.8 cm < x < 1.0 cm to nearest mm. [1] (ii) Absolute uncertainty 1 or 2 mm (or half the range of repeats), and correct method

of calculation. [1] (c) (ii) Measurement for h to nearest mm. [1] (d) (iii) Value for t > 1 s and given to 0.1 s or 0.01 s. [1] Check raw data if there are repeats. (iv) Correct calculation of R, with consistent unit (e.g. cm3 s–1). [1] (e) (i) Values for x, V and h. [1] (ii) Correct trend (R increases with h). [1] (f) (i) Values of k calculated correctly. [1] (ii) Valid conclusion based on the calculated values of k. Candidate must test against

a stated criterion. [1]



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(g)

(i) Problems 4 max (ii) Improvements 4 max No credit/not enough

A Two readings are not enough (to draw a conclusion).

Take more readings, and plot a graph/calculate more k values.

More readings and calculate the average/ only one reading.

B Bottle not circular/ diameter at P different to that at Q.

Collect water and measure volume/remeasure diameter at P.

C Bottle deforms when measuring d.

Use vernier callipers to measure d.

Use string to measure d.

D Difficult to see water level/meniscus problems/refraction problems.

Use coloured water/liquid. Use oil.

E Labels get wet/ink runs Use waterproof labels/ink

F Difficult to judge when to start/stop timing.

Use video, with timing method.

Human reaction time error.

G Large uncertainty in x. Use travelling microscope to measure x.

X Another valid point E.g. Flowrate calculated is not the flowrate at h.

E.g. Measure h to point midway between marks.

Move marks closer together.

Ignore ‘parallax problems’ unless there is a convincing diagram. Ignore ‘use assistant’. Ignore ‘use distance sensor’ unless there is a convincing diagram. Ignore ‘use a computer/datalogger/light gates’. Ignore ‘bottle not vertical’. [Total: 20]





9702 PHYSICS







© UCLES 2010

1 (a) (i) Value of d to the nearest 0.01 mm or 0.001 mm with consistent unit. [1] 0.20 < d < 0.60 mm.

(b) (iii) Value of x in range 40 cm–60 cm with consistent unit. [1]

Value of I with units.

(c) Six sets of readings of x and I scores 5 marks, five sets scores 4 marks etc. Incorrect trend then –1. Minor help from supervisor –1 ; major help from supervisor –2 [5] Range [1] xmax > 70 cm; xmin < 30 cm Column headings [1] Each column heading must contain a quantity and a unit. There must be some distinguishing mark between the quantity and the unit

(solidus is expected but accept, for example, 1/I (A–1). Do not allow 1/I (A)) Consistency of presentation of raw readings. [1] All values of x must be given to the nearest mm. Significant figures [1]

S.F. in 1/I must be the same as, or one more than, the least number of significant

figures used in raw I. Calculation [1]

Correct calculation of 1/I.



© UCLES 2010

(d) (i) Axes [1] Sensible scales must be used. Awkward scales (e.g. 3:10) are not allowed. Scales must be chosen so that the plotted points occupy at least half the graph grid in both x and y directions. Scales must be labelled with the quantity which is being plotted. Ignore units. Scale markings should be no more than three large squares apart. Plotting of points [1] All observations must be plotted on the grid. Do not accept blobs (points with diameter > 0.5 small square). Ring and check a suspect plot. Work to an accuracy of half a small square.

(ii) Line of best fit [1]

Judge by the balance of at least 5 points about the candidate’s line. There must be an even distribution of points either side of the line along the full length. Lines must not be kinked. Do not accept lines thicker than half a small square. Quality [1] All points in the table (minimum 5) must be plotted for this mark to be scored. All points must be within 2 cm (to scale) in x direction of a straight line.

(iii) Gradient [1]

The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square. Intercept [1] Either: Check correct read-off from a point on the line, and substitution into y = mx + c. Read-off must be accurate to half a small square. Allow ecf of gradient value. Or: Check read-off of intercept directly from graph.

(e) Values obtained in (a)(ii) and (d)(iii) substituted correctly into equation: ARN

M ρ= [1]

Do not allow substitution methods to find M or N

Value for ρ in range: 1 × 10–7 Ω m – 5 × 10–6 Ω m with consistent unit. [1] [Total: 20]



© UCLES 2010

2 (a) (ii) Measurement of x to nearest mm. x < 15.0 cm with consistent unit. [1] –1 for supervisor’s help.

(b) (iii) Measurement of θ (less than 90°) with unit. [1] (iv) Absolute uncertainty in θ in the range 2°–10°. [1]

If repeated readings have been taken, then the uncertainty can be half the range. Correct method of calculation of percentage uncertainty.

(v) m = 50 g with consistent unit [1]

M = 60 g with consistent unit [1]

(vi) Correct calculation of m/M (0.83 or 0.833). No units. [1] (c) Measurement of θ [1]

m = 40 g; M = 70 g [1] Quality: θ2 > θ1 [1]

(d) (i) Correct calculation of two values of k. [1] (ii) Justification of sf in k linked to θ, m and M [1] (iii) Valid conclusion based on the calculated values of k. [1]




© UCLES 2010

(e) Identifying limitations (4 marks) and suggesting improvements (4 marks)


A Two readings are not enough (to draw a conclusion.

Take more readings and plot a graph/calculate more k values (and compare).

Few readings. Take more readings and calculate average. Only one reading.

B Difficult to balance with reason e.g. unstable or effect of fans/draughts/a.c.

Drill hole higher up/switch off fans/a.c./close windows.

Closed room.

C Difficult to judge when wooden strip horizontal/parallel (to the bench).

Method of ensuring strip is horizontal/parallel to bench e.g. use a spirit level or metre rule(s) to measure height of both ends/sight against window. Allow detailed use of set square.

Strip not straight/parallel/ horizontal. Use set square.

D Difficult keeping x constant/ weights move.

Method of fixing cotton loop to rule e.g. tape, glue.

E Difficult to measure θ because hard to judge vertical/movement of hand.

Use a plumb line/clamped ruler/clamp protractor.

Bigger protractor. Paper behind protractor.

F Friction at pulley/between nail and wooden strip.

Use lubricant/method of reducing friction.

Friction. Better pulley/ smooth(er) string/thin(ner) string. Friction between string and pulley. Lubrication between string and pulley.

G Mass (values) not accurate. Use balance/method of weighing mass.

Weigh masses.

Do not credit ‘parallax problems’, ‘use assistant’ or references to sensors, computers or

videos. [Total: 20]





9702 PHYSICS







© UCLES 2010

1 (c) Measurements for all raw l in range 19.5 to 20.5 cm. [1] (e) (i) Measurements for all raw h1 and h2 to nearest mm. [1] (iii) Measurement for raw d to nearest mm, with unit, in range 1.5 to 2.5 cm. [1] (f) Five sets of readings of h1, h2 and d scores 4 marks, four sets scores 3 marks etc. [4] Incorrect trend then –1. Help from supervisor then –1. Range – [1] d values used must include dmin ≤ 3 cm and dmax ≥ 8 cm Column headings – [1] Each column heading must contain a quantity and a unit where appropriate. There must be some distinguishing mark between the quantity and the unit.

e.g. θ /o , 1/tan θ, sin

θ, sin (θ /o) not sin θ /o, not (1/tan

θ)/o Consistency of presentation of raw readings – [1] All h values in the table must be given to the same precision. Significant figures – [1]

S.f. for 1/tan θ must be the same as, or one more than, the minimum s.f. given

for (h1 – h2) and l. Calculation – [1]

1/tan θ calculated correctly.

(Graph) Axes – [1] Sensible scales must be used, no awkward scales (e.g. 3:10). Scales must be chosen so that the plotted points must occupy at least half the graph grid in both x and y directions. Scales must be correctly labelled with the quantity that is being plotted. Ignore units. Scale markings must be no more than three large squares apart. Plots – [1] All observations must be plotted. Ring and check a suspect plot. Tick if correct. Re-plot if incorrect. Work to an accuracy of half a small square. Diameter of plots must be ≤ half a small square (no blobs). Line of best fit – [1] Judge by balance of all plots, at least 4 trend points, about the candidate's line. There must be an even distribution of points either side of the line along the full length. Line must not be kinked. Quality – [1] Scatter of points must be less than ± 0.25 cm in the d direction about the examiner’s line. All points in table must be plotted (at least 4) for this mark to be scored.



© UCLES 2010

(g) (iii) Gradient [1] The hypotenuse of the triangle must be at least half the length of the drawn line. Both read-offs must be accurate to half a small square. Intercept [1] Check that the read-off from graph or the method of calculation (substitution of correct read offs into y = mx + c) is correct. (h) Value of a = gradient and value of b = intercept. [1] Unit for a (m–1 cm–1 or mm–1) consistent with value and b (no unit). [1] [Total: 20] 2 (b) (i) Raw length and width to nearest mm with unit. Help from supervisor –1 [1] Values of length and width in range 1 cm to 10 cm. [1] Correct calculation of A, with consistent unit. [1] (ii) S.f. in A same as/one more than the (smallest) s.f. in length and width [1] (not just “raw readings”). (d) (i) Measurement of F, with unit, F < 10 N. [1] Evidence of repeated measurements of F. [1] (ii) Uncertainty in measurements of F stated, in range 0.1 to 0.5 N. [1] (e) Values of second length and second width. [1] Correct calculation of A. [1] Measurement of F. [1] Second F = first F (within 1 N). [1] (f) Justification of a valid conclusion based on two values of F being within (or outside) the uncertainty in (d)(ii). [1]



© UCLES 2010

(g)

(i) Limitations 4 max (ii) Improvements 4 max No credit/not enough

A Two readings not enough (to draw a conclusion)/too few readings/only two readings.

Take many readings for different areas and plot a graph/compare more F values

Repeat readings Few readings One reading NOT average F

B Maximum force reached without warning(suddenly)/ reading over quickly, link to short time

Method of recording maximum reading e.g. force sensor + data logger/video recording to find force/meter which retains max reading/ use masses and pulley system

Position sensors /parallax/computer methods/bald human reaction time error/ increase force slowly/fast paper/high speed camera/ slow camera

C Reason for the problem of detecting paper movement/ difficult to look at meter and paper at same time.

Method to indicate movement e.g. contrasting colours of paper/drawing a reference mark

Difficult to know when paper moves. Fast movement

D Position of eraser (and weights) not fixed/ Mass(weight) of eraser changes/irregularity of rubber shape (not rectangular)

Method to ensure same position e.g. mark position on top paper/method to ensure constant mass e.g. use malleable strip which can be bent to change A/ change total masses to account for change in mass of rubber/pile up unused rubber pieces on top/improved method to measure rubber e.g. vernier caliper

Keep mass constant

E Variation in direction of force/misalignment of paper strips (which affects F).

Method to ensure direction is constant e.g. align strips along straight edge/draw a line to follow/method to equalise levels

F Uneven bench surface (leading to contact area being less than A).

Method to ensure smoother surface e.g. use named surface e.g. glass or melamine/sand the surface

Use smoother surface

X: Increase mass so increase the force (reducing % uncertainty in force). Do not credit references to zero error/accuracy/digital meter friction between papers/re-zeroing after each experiment/2 people/paper tearing/clip deforming.

[Total: 20]





9702 PHYSICS

9702/41 Paper 4 (A2 Structured Questions), maximum raw mark 100






© UCLES 2010

Section A 1 (a) force per unit mass (ratio idea essential) B1 [1] (b) graph: correct curvature M1 from (R,1.0 gS) & at least one other correct point A1 [2] (c) (i) fields of Earth and Moon are in opposite directions M1 either resultant field found by subtraction of the field strength or any other sensible comment A1 so there is a point where it is zero A0 [2] (allow FE = –FM for 2 marks) (ii) GME / x

2 = GMM / (D – x)2 C1 (6.0 × 1024) / (7.4 × 1022) = x2 / (60RE – x)2 C1 x = 54 RE A1 [3] (iii) graph: g = 0 at least ⅔ distance to Moon B1 gE and gM in opposite directions M1 correct curvature (by eye) and gE > gM at surface A1 [3] 2 (a) (i) no forces (of attraction or repulsion) between atoms / molecules / particles B1 [1] (ii) sum of kinetic and potential energy of atoms / molecules M1 due to random motion A1 [2] (iii) (random) kinetic energy increases with temperature M1 no potential energy (so increase in temperature increases internal energy) A1 [2] (b) (i) zero A1 [1]

(ii) work done = p∆V C1

= 4.0 × 105 × 6 × 10–4 = 240 J (ignore any sign) A1 [2] (iii)

change work done / J heating / J increase in internal energy / J

P → Q Q → R R → P

240 0

–560

0 320

0

240 320

–560

(correct signs essential) (each horizontal line correct, 1 mark – max 3) B3 [3]



© UCLES 2010

3 (a) (i) resonance B1 [1] (ii) amplitude 16 mm and frequency 4.6 Hz A1 [1]

(b) (i) a = (–)ω2x and ω = 2πf C1

a = 4π2 × 4.62 × 16 × 10–3 C1 = 13.4 m s–2 A1 [3] (ii) F = ma C1 = 50 × 10–3 × 13.4 = 2.0 N A1 [2] (c) line always ‘below’ given line and never zero M1 peak is at 4.6 Hz (or slightly less) and flatter A1 [2] 4 (a) charge / potential (difference) (ratio must be clear) B1 [1]

(b) (i) V = Q / 4πε0r B1 [1]

(ii) C = Q / V = 4πε0r and 4πε0 is constant M1 so C ~ r A0 [1]

(c) (i) r = C / 4πε0r C1

r = (6.8 × 10–12) / (4π × 8.85 × 10–12) C1 = 6.1 × 10–2

m A1 [3]

(ii) Q = CV = 6.8 × 10–12 × 220

= 1.5 × 10–9 C A1 [1]

(d) (i) V = Q/C = (1.5 × 10–9) / (18 × 10–12) = 83 V A1 [1] (ii) either energy = ½CV 2 C1

∆E = ½ × 6.8 × 10–12 × 2202 – ½ × 18 × 10–12 × 832 C1 = 1.65 × 10–7 – 6.2 × 10–8 = 1.03 × 10–7

J A1 [3] or energy = ½QV (C1)

∆E = ½ × 1.5 × 10–9 × 220 – ½ × 1.5 × 10–9 × 83 (C1) = 1.03 × 10–7

J (A1)



© UCLES 2010

5 (a) field into (the plane of) the paper B1 [1]

(b) force due to magnetic field provides the centripetal force B1 mv2 / r = Bqv C1 B = (20 × 1.66 × 10–27 × 1.40 × 105) / (1.6) × 10–19 × 6.4 × 10–2) B1 = 0.454 T A0 [3]

(c) (i) semicircle with diameter greater than 12.8 cm B1 [1]

(ii) new flux density = 20

22 × 0.454 C1

B = 0.499 T A1 [2]

6 (a) (i) e.g. prevent flux losses / improve flux linkage B1 [1] (ii) flux in core is changing B1 e.m.f. / current (induced) in core B1 induced current in core causes heating B1 [3]

(b) (i) that value of the direct current producing same (mean) power / heating M1 in a resistor A1 [2] (ii) power in primary = power in secondary M1

VP IP = VS IS A1 [2]

7 (a) (i) e.g. electron / particle diffraction B1 [1] (ii) e.g. photoelectric effect B1 [1]

(b) (i) 6 A1 [1] (ii) change in energy = 4.57 × 10–19 J

λ = hc / E C1

= (6.63 × 10–34 × 3.0 × 108) / (4.57 × 10–19)

= 4.4 × 10–7 m A1 [2]

8 (a) splitting of a heavy nucleus (not atom/nuclide) M1 into two (lighter) nuclei of approximately same mass A1 [2]

(b) n1

0

He4

2 (allow α

4

2) M2

Li7

3 A1 [3]

(c) emitted particles have kinetic energy B1 range of particles in the control rods is short / particles stopped in rods / lose kinetic energy in rods B1 kinetic energy of particles converted to thermal energy B1 [3]



© UCLES 2010

Section B 9 (a) (i) non-inverting (amplifier) B1 [1] (ii) (G =) 1 + R2 / R1 B1 [1]

(b) (i) gain = 1 + 100 / 820 C1 output = 17 mV A1 [2] (ii) 9 V A1 [1] (R2 / R1 scores 0 in (a)(ii) but possible 1 mark in each of (b)(i) and (b)(ii) (1 + R1 / R2) scores 0 in (a)(ii), no mark in (b)(i), possible 1 mark in (b)(ii) (1 – R2 / R1) or R1 / R2 scores 0 in (a)(ii), (b)(i) and (b)(ii)) 10 (a) (i) density × speed of wave (in the medium) B1 [1]

(ii) ρ = (7.0 × 106) / 4100 = 1700 kg m–3 A1 [1]

(b) (i) I = IT + IR B1 [1]

(ii) 1. α = (0.1 × 106)2 / (3.1 × 106)2 C1 = 0.001 A1 [2]

2. α ≈ 1 A1 [1]

(c) either very little transmission at an air-skin boundary M1 (almost) complete transmission at a gel-skin boundary M1 when wave travels in or out of the body A1 [3] or no gel, majority reflection (M1) with gel, little reflection (M1) when wave travels in or out of the body (A1)

11 (a) (i) unwanted random power / signal / energy B1 [1] (ii) loss of (signal) power / energy B1 [1] (b) (i) either signal-to-noise ratio at mic. = 10 lg (P2 / P1) C1 = 10 lg (2.9 × 10–6 / 3.4 × 10–9) = 29 dB A1 maximum length = (29 – 24) / 12 C1 = 0.42 km = 420 m A1 [4] or signal-to-noise ratio at receiver = 10 lg (P2 / P1) (C1) at receiver, 24 = 10 lg(P / 3.4 × 10–9) P = 8.54 × 10–7 W (A1) power loss in cables = 10 lg(2.9 × 10–6 / 8.54 × 10–7) (C1) = 5.3 dB length = 5.3 / 12 km = 440 m (A1)



© UCLES 2010

(ii) use an amplifier M1 coupled to the microphone A1 [2] (repeater amplifiers scores no mark) 12 (a) (carrier wave) transmitted from Earth to satellite (1) satellite receives greatly attenuated signal (1) signal amplified and transmitted back to Earth B1 at a different (carrier) frequency B1 different frequencies prevent swamping of uplink signal (1) e.g. of frequencies used (6/4 GHz, 14/11 GHz, 30/20 GHz) (1) (two B1 marks plus any two other for additional physics) B2 [4] (b) advantage: e.g. much shorter time delay M1 because orbits are much lower A1 e.g. whole Earth may be covered (M1) in several orbits / with network (A1) disadvantage: e.g. either must be tracked or limited use in any one orbit M1 more satellites required for continuous operation A1 [4]





9702 PHYSICS







© UCLES 2010

Section A 1 (a) force per unit mass (ratio idea essential) B1 [1] (b) graph: correct curvature M1 from (R,1.0 gS) & at least one other correct point A1 [2] (c) (i) fields of Earth and Moon are in opposite directions M1 either resultant field found by subtraction of the field strength or any other sensible comment A1 so there is a point where it is zero A0 [2] (allow FE = –FM for 2 marks) (ii) GME / x

2 = GMM / (D – x)2 C1 (6.0 × 1024) / (7.4 × 1022) = x2 / (60RE – x)2 C1 x = 54 RE A1 [3] (iii) graph: g = 0 at least ⅔ distance to Moon B1 gE and gM in opposite directions M1 correct curvature (by eye) and gE > gM at surface A1 [3] 2 (a) (i) no forces (of attraction or repulsion) between atoms / molecules / particles B1 [1] (ii) sum of kinetic and potential energy of atoms / molecules M1 due to random motion A1 [2] (iii) (random) kinetic energy increases with temperature M1 no potential energy (so increase in temperature increases internal energy) A1 [2] (b) (i) zero A1 [1]

(ii) work done = p∆V C1

= 4.0 × 105 × 6 × 10–4 = 240 J (ignore any sign) A1 [2] (iii)

change work done / J heating / J increase in internal energy / J

P → Q Q → R R → P

240 0

–560

0 320

0

240 320

–560

(correct signs essential) (each horizontal line correct, 1 mark – max 3) B3 [3]



© UCLES 2010

3 (a) (i) resonance B1 [1] (ii) amplitude 16 mm and frequency 4.6 Hz A1 [1]

(b) (i) a = (–)ω2x and ω = 2πf C1

a = 4π2 × 4.62 × 16 × 10–3 C1 = 13.4 m s–2 A1 [3] (ii) F = ma C1 = 50 × 10–3 × 13.4 = 2.0 N A1 [2] (c) line always ‘below’ given line and never zero M1 peak is at 4.6 Hz (or slightly less) and flatter A1 [2] 4 (a) charge / potential (difference) (ratio must be clear) B1 [1]

(b) (i) V = Q / 4πε0r B1 [1]

(ii) C = Q / V = 4πε0r and 4πε0 is constant M1 so C ~ r A0 [1]

(c) (i) r = C / 4πε0r C1

r = (6.8 × 10–12) / (4π × 8.85 × 10–12) C1 = 6.1 × 10–2

m A1 [3]

(ii) Q = CV = 6.8 × 10–12 × 220

= 1.5 × 10–9 C A1 [1]

(d) (i) V = Q/C = (1.5 × 10–9) / (18 × 10–12) = 83 V A1 [1] (ii) either energy = ½CV 2 C1

∆E = ½ × 6.8 × 10–12 × 2202 – ½ × 18 × 10–12 × 832 C1 = 1.65 × 10–7 – 6.2 × 10–8 = 1.03 × 10–7

J A1 [3] or energy = ½QV (C1)

∆E = ½ × 1.5 × 10–9 × 220 – ½ × 1.5 × 10–9 × 83 (C1) = 1.03 × 10–7

J (A1)



© UCLES 2010

5 (a) field into (the plane of) the paper B1 [1]

(b) force due to magnetic field provides the centripetal force B1 mv2 / r = Bqv C1 B = (20 × 1.66 × 10–27 × 1.40 × 105) / (1.6) × 10–19 × 6.4 × 10–2) B1 = 0.454 T A0 [3]

(c) (i) semicircle with diameter greater than 12.8 cm B1 [1]

(ii) new flux density = 20

22 × 0.454 C1

B = 0.499 T A1 [2]

6 (a) (i) e.g. prevent flux losses / improve flux linkage B1 [1] (ii) flux in core is changing B1 e.m.f. / current (induced) in core B1 induced current in core causes heating B1 [3]

(b) (i) that value of the direct current producing same (mean) power / heating M1 in a resistor A1 [2] (ii) power in primary = power in secondary M1

VP IP = VS IS A1 [2]

7 (a) (i) e.g. electron / particle diffraction B1 [1] (ii) e.g. photoelectric effect B1 [1]

(b) (i) 6 A1 [1] (ii) change in energy = 4.57 × 10–19 J

λ = hc / E C1

= (6.63 × 10–34 × 3.0 × 108) / (4.57 × 10–19)

= 4.4 × 10–7 m A1 [2]

8 (a) splitting of a heavy nucleus (not atom/nuclide) M1 into two (lighter) nuclei of approximately same mass A1 [2]

(b) n1

0

He4

2 (allow α

4

2) M2

Li7

3 A1 [3]

(c) emitted particles have kinetic energy B1 range of particles in the control rods is short / particles stopped in rods / lose kinetic energy in rods B1 kinetic energy of particles converted to thermal energy B1 [3]



© UCLES 2010

Section B 9 (a) (i) non-inverting (amplifier) B1 [1] (ii) (G =) 1 + R2 / R1 B1 [1]

(b) (i) gain = 1 + 100 / 820 C1 output = 17 mV A1 [2] (ii) 9 V A1 [1] (R2 / R1 scores 0 in (a)(ii) but possible 1 mark in each of (b)(i) and (b)(ii) (1 + R1 / R2) scores 0 in (a)(ii), no mark in (b)(i), possible 1 mark in (b)(ii) (1 – R2 / R1) or R1 / R2 scores 0 in (a)(ii), (b)(i) and (b)(ii)) 10 (a) (i) density × speed of wave (in the medium) B1 [1]

(ii) ρ = (7.0 × 106) / 4100 = 1700 kg m–3 A1 [1]

(b) (i) I = IT + IR B1 [1]

(ii) 1. α = (0.1 × 106)2 / (3.1 × 106)2 C1 = 0.001 A1 [2]

2. α ≈ 1 A1 [1]

(c) either very little transmission at an air-skin boundary M1 (almost) complete transmission at a gel-skin boundary M1 when wave travels in or out of the body A1 [3] or no gel, majority reflection (M1) with gel, little reflection (M1) when wave travels in or out of the body (A1)

11 (a) (i) unwanted random power / signal / energy B1 [1] (ii) loss of (signal) power / energy B1 [1] (b) (i) either signal-to-noise ratio at mic. = 10 lg (P2 / P1) C1 = 10 lg (2.9 × 10–6 / 3.4 × 10–9) = 29 dB A1 maximum length = (29 – 24) / 12 C1 = 0.42 km = 420 m A1 [4] or signal-to-noise ratio at receiver = 10 lg (P2 / P1) (C1) at receiver, 24 = 10 lg(P / 3.4 × 10–9) P = 8.54 × 10–7 W (A1) power loss in cables = 10 lg(2.9 × 10–6 / 8.54 × 10–7) (C1) = 5.3 dB length = 5.3 / 12 km = 440 m (A1)



© UCLES 2010

(ii) use an amplifier M1 coupled to the microphone A1 [2] (repeater amplifiers scores no mark) 12 (a) (carrier wave) transmitted from Earth to satellite (1) satellite receives greatly attenuated signal (1) signal amplified and transmitted back to Earth B1 at a different (carrier) frequency B1 different frequencies prevent swamping of uplink signal (1) e.g. of frequencies used (6/4 GHz, 14/11 GHz, 30/20 GHz) (1) (two B1 marks plus any two other for additional physics) B2 [4] (b) advantage: e.g. much shorter time delay M1 because orbits are much lower A1 e.g. whole Earth may be covered (M1) in several orbits / with network (A1) disadvantage: e.g. either must be tracked or limited use in any one orbit M1 more satellites required for continuous operation A1 [4]





9702 PHYSICS







© UCLES 2010

Section A 1 (a) (i) rate of change of angle / angular displacement M1 swept out by radius A1 [2] (ii) ω × T = 2π B1 [1] (b) centripetal force is provided by the gravitational force B1 either mr(2π/T)2 = GMm/r 2 or mrω 2 = GMm/r 2 M1 r 3 × 4π2 = GM × T 2 A1 GM/4π2 is a constant (c) A1 T 2 = cr 3 A0 [4] (c) (i) either T 2 = (45/1.08)3 × 0.6152 or T 2 = 0.30 × 453 C1 T = 165 years A1 [2] (ii) speed = (2π × 1.08 × 108) / (0.615 × 365 × 24 × 3600) C1 = 35 km s–1 A1 [2] 2 (a) atoms / molecules / particles behave as elastic (identical) spheres (1) volume of atoms / molecules negligible compared to volume of containing vessel (1) time of collision negligible to time between collisions (1) no forces of attraction or repulsion between atoms / molecules (1) atoms / molecules / particles are in (continuous) random motion (1) (any four, 1 each) B4 [4] (b) pV =

3

1 Nm<c2> and pV = nRT or pV = NkT B1

3

1 Nm<c2> = nRT or = NkT and <EK> = ½m<c2> B1

n = N/NA or k = R/NA B1

<EK> = 2

3 × R/NA × T A0 [3]

(c) (i) reaction represents either build-up of nucleus from light nuclei or build-up of heavy nucleus from nuclei M1 so fusion reaction A1 [2] (ii) proton and deuterium nucleus will have equal kinetic energies B1 1.2 × 10–14 =

2

3 × 8.31 / (6.02 × 1023) × T C1

T = 5.8 × 108 K A1 [3] (use of E = 2.4 × 10–14 giving 1.16 × 109 K scores 1 mark) (iii) either inter-molecular / atomic / nuclear forces exist or proton and deuterium nucleus are positively charged / repel B1 [1]



© UCLES 2010

3 (a) (i) 8.0 cm A1 [1] (ii) 2πf = 220 C1 f = 35 (condone unit) A1 [2] (iii) line drawn mid-way between AB and CD (allow ±2 mm) B1 [1] (iv) v = ωa C1 = 220 × 4.0 = 880 cm s–1 A1 [2] (b) (i) 1. line drawn 2 cm above AB (allow ±2 mm) B1 [1] 2. arrow pointing upwards B1 [1] (ii) 1. line drawn 2 cm above AB (allow ±2 mm) B1 [1] 2. arrow pointing downwards B1 [1] (iii) v = ω√(a2 – x2) = 220 × √(4.02 – 2.02) C1 = 760 cm s–1 A1 [2] (incorrect value for x, 0/2 marks) 4 (a) (i) work done moving unit positive charge M1 from infinity to the point A1 [2] (ii) charge / potential (difference) (ratio must be clear) B1 [1] (b) (i) capacitance = (2.7 × 10–6) / (150 × 103) C1 (allow any appropriate values) capacitance = 1.8 × 10–11 (allow 1.8 ±0.05) A1 [2] (ii) either energy = ½CV 2 or energy = ½QV and Q = CV C1 energy = ½ × 1.8 × 10–11 × (150 × 103)2 or ½ × 2.7 × 10–6 × 150 × 103 = 0.20 J A1 [2] (c) either since energy ~ V 2, capacitor has (½)2 of its energy left or full formula treatment C1 energy lost = 0.15 J A1 [2]



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5 (a) magnetic flux = BA = 89 × 10–3 × 5.0 × 10–2 × 2.4 × 10–2 C1 = 1.07 × 10–4 Wb A1 [2] (b) (i) e.m.f. = ∆φ / ∆t C1 (for ∆φ = 1.07 × 10–4 Wb), ∆t = 2.4 × 10–2 / 1.8 = 1.33 × 10–2 s C1 e.m.f. = (1.07 × 10–4) / (1.33 × 10–2) = 8.0 × 10–3 V A1 [3] (ii) current = 8.0 × 10–3 / 0.12 M1 ≈ 70 mA A0 [1] (c) force on wire = BIL = 89 × 10–3 × 70 × 10–3 × 6.0 × 10–2 C1 ≈ 4 × 10–4 (N) M1 suitable comment e.g. this force is too / very small (to be felt) A1 [3] 6 (a) power / heating depends on I2 M1 so independent of current direction A1 [2] (b) either maximum power = I0

2R or average power = IRMS2R M1

I0 = √2 × IRMS M1 maximum power = 2 × average power ratio = 0.5 A1 [3] 7 (a) force due to E-field is equal and opposite to force due to B-field B1 Eq = Bqv B1 v = E/B B1 [3] (b) either charge and mass are not involved in the equation in (a) or FE and FB are both doubled or E, B and v do not change M1 so no deviation A1 [2] 8 (a) minimum frequency for electron to be emitted (from surface) M1 of electromagnetic radiation / light / photons A1 [2] (b) E = hc / λ or E = hf and c = fλ C1 either threshold wavelength = (6.63 × 10–34 × 3.0 × 108) / (5.8 × 10–19) = 340 nm or energy of 340 nm photon = 4.4 × 10–19 J or threshold frequency = 8.7 × 1014 Hz or 450 nm → 6.7 × 1014 Hz A1 appropriate comment comparing wavelengths / energies / frequencires B1 so no effect on photo-electric current B1 [4]



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Section B 9 (a) (i) edges can be (clearly) distinguished B1 [1] (ii) e.g. size of X-ray source / anode / target / aperture scattering of X-ray beam pixel size (any two, 1 each) B2 further detail e.g. use of lead grid B1 [3] (b) X-ray image involves a single exposure B1 CT scan: exposure of a slice from many different angles M1 repeated for different slices A1 CT scan involves a (much) greater exposure B1 [4] 10 (a) e.g. infinite input impedance / resistance zero output impedance / resistance infinite gain infinite bandwidth infinite slew rate (any three, 1 each) B3 [3] (b) (i) with switch open, V – is less (positive) than V + M1 output is negative A1 with switch closed, V – is more (positive) than V + so output is positive A1 [3] (allow similar scheme if V – more positive than V + treated first) (ii) 1. diodes connected correctly between output and earth M1 2. green identified correctly A1 [2] (do not allow this mark if not argued in (i)) 11 (a) (i) I / I0 = exp(–1.5 × 2.9) C1 = 0.013 A1 [2] (ii) I / I0 = exp(–4.6 × 0.95) = 0.013 A1 [1]

(b) attenuation (coefficients) in muscle and in fat are similar B1 attenuation (coefficients) in bone and muscle / fat are different B1 contrast depends on difference in attenuation B1 [3]



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12 (a) (i) 1. signal has same variation (with time) as the data B1 2. consists of (a series of) ‘highs’ and ‘lows’ B1 either analogue is continuously variable (between limits) or digital has no intermediate values B1 [3] (ii) e.g. can be regenerated / noise can be eliminated extra data can be added to check / correct transmitted signal (any two reasonable suggestions, 1 each) B2 [2] (b) (i) analogue signal is sampled at (regular time) intervals B1 sampled signal is converted into a binary number B1 [2] (ii) one channel is required for each bit (of the digital number) B1 [1]





9702 PHYSICS

9702/51 Paper 5 (Planning, Analysis and Evaluation), maximum raw mark 30






© UCLES 2010

1 Planning (15 marks) Defining the problem (3 marks) P1 f is the independent variable and V is the dependent variable or vary f and measure V [1] P2 Keep the current in coil X constant [1] P3 Keep the number of turns on coil (Y)/area of coil Y constant Do not credit reference to coil X only. [1] Methods of data collection (5 marks) M1 Two independent coils labelled X and Y. [1] M2 Alternating power supply/signal generator connected to coil X in a workable circuit. [1] M3 Coil Y connected to voltmeter/c.r.o. in a workable circuit. [1] M4 Use c.r.o. to determine period/frequency or read off signal generator. [1] M5 Method to keep current constant in coil X: adjust signal generator/use of rheostat. [1] Method of analysis (2 marks) A1 Plot a graph of V against f. [1] A2 Relationship valid if straight line through origin [1] Safety considerations (1 mark) S1 Reference to hot coils – switch off when not in use/use gloves/do not touch coils. Must refer

to hot coils. [1] Additional detail (4 marks) D1/2/3/4 Relevant points might include [4]

1. Use large current in coil X/large number of coils on coil Y (to increase emf). 2. Use iron core (to increase emf).

3. Detail on measuring emf e.g. height × y-gain. 4. Avoid other alternating magnetic fields. 5. Detail on measuring frequency from c.r.o. to determine period and hence f. 6. Use of ammeter/c.r.o. and resistor to check current is constant 7. Use insulated wire for coils. 8. Keep coil Y and coil X in the same relative positions.

Do not allow vague computer methods.

[Total: 15]



© UCLES 2010

2 Analysis, conclusions and evaluation (15 marks)

Part Mark Expected Answer Additional Guidance

(a) A1 Gradient = b y-intercept = lg a

Allow log a but not ln a

(b) T1 T2

1.9777 0.292 or 0.2923

1.9294 0.265 or 0.2648

1.8751 0.241 or 0.2405

1.8129 0.210 or 0.2095

1.7404 0.170 or 0.1703

1.6532 0.127 or 0.1271

T1 for lg l column – ignore rounding errors; min 2 dp. T2 for lg T column – must be values given A mixture is allowed

U1 From ± 0.004 or ± 0.005 to ± 0.006

or ± 0.007

Allow more than one significant figure.

(c) (i) G1 Six points plotted correctly Must be within half a small square; penalise ≥

half a small square. Penalise ‘blobs’ ≥ half a small square. Ecf allowed from table.

U2 Error bars in lg (T/s) plotted correctly.

All error bars must be plotted. Check first and last point. Must be accurate within half a small

square; penalise ≥ half a small square.

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1.65, 0.124) and (1.65, 0.128) and upper end of line should pass between (2.00, 0.300) and (2.00, 0.306). Allow ecf from points plotted incorrectly; five trend plots needed – examiner judgement.

G3 Worst acceptable straight line. Steepest or shallowest possible line that passes through all the error bars.

Line should be clearly labelled or dashed. Should pass from top of top error bar to bottom of bottom error bar or bottom of top error bar to top of bottom error bar. Mark scored only if all error bars are plotted.

(iii) C1 Gradient of best fit line

The triangle used should be at least half the length of the drawn line. Check the read offs.

Work to half a small square; penalise ≥ half a small square.

U3 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient and gradient.

(iv) C2 y-intercept Must be negative. Check substitution of point from line into c = y – mx. Allow ecf from (c)(iii).



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U4 Uncertainty in y-intercept Method of determining absolute uncertainty Difference in worst y-intercept and y-intercept. Do not allow ecf from false origin read-off (FOX). Allow ecf from (c)(iv).

(d) C3 a = 10y-intercept y-intercept must be used. Expect an answer of about 0.19. If FOX expect answer of about 1.3.

C4 b = gradient and in the range 0.495 to 0.520 and to 2 or 3 sf

Allow 0.50 to 0.52 to 2 sf Penalise 1 sf or ≥4 sf

U5 Absolute uncertainty in a and b Difference in a and worst a. Uncertainty in b should be the same as the uncertainty in the gradient.

[Total: 15]

Uncertainties in Question 2 (c) (iii) Gradient [U3]

1. Uncertainty = gradient of line of best fit – gradient of worst acceptable line

2. Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (c) (iv) [U4]

1. Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line

2. Uncertainty = ½ (y-intercept of steepest worst line – y-intercept of shallowest worst line)

(d) [U5]

1. Uncertainty = 10 best y-intercept - 10 worst y-intercept





9702 PHYSICS







© UCLES 2010

1 Planning (15 marks) Defining the problem (3 marks) P1 f is the independent variable and V is the dependent variable or vary f and measure V [1] P2 Keep the current in coil X constant [1] P3 Keep the number of turns on coil (Y)/area of coil Y constant Do not credit reference to coil X only. [1] Methods of data collection (5 marks) M1 Two independent coils labelled X and Y. [1] M2 Alternating power supply/signal generator connected to coil X in a workable circuit. [1] M3 Coil Y connected to voltmeter/c.r.o. in a workable circuit. [1] M4 Use c.r.o. to determine period/frequency or read off signal generator. [1] M5 Method to keep current constant in coil X: adjust signal generator/use of rheostat. [1] Method of analysis (2 marks) A1 Plot a graph of V against f. [1] A2 Relationship valid if straight line through origin [1] Safety considerations (1 mark) S1 Reference to hot coils – switch off when not in use/use gloves/do not touch coils. Must refer

to hot coils. [1] Additional detail (4 marks) D1/2/3/4 Relevant points might include [4]

1. Use large current in coil X/large number of coils on coil Y (to increase emf). 2. Use iron core (to increase emf).

3. Detail on measuring emf e.g. height × y-gain. 4. Avoid other alternating magnetic fields. 5. Detail on measuring frequency from c.r.o. to determine period and hence f. 6. Use of ammeter/c.r.o. and resistor to check current is constant 7. Use insulated wire for coils. 8. Keep coil Y and coil X in the same relative positions.


[Total: 15]



© UCLES 2010



(a) A1 Gradient = b y-intercept = lg a

Allow log a but not ln a

(b) T1 T2

1.9777 0.292 or 0.2923

1.9294 0.265 or 0.2648

1.8751 0.241 or 0.2405

1.8129 0.210 or 0.2095

1.7404 0.170 or 0.1703

1.6532 0.127 or 0.1271

T1 for lg l column – ignore rounding errors; min 2 dp. T2 for lg T column – must be values given A mixture is allowed

U1 From ± 0.004 or ± 0.005 to ± 0.006

or ± 0.007


(c) (i) G1 Six points plotted correctly Must be within half a small square; penalise ≥

half a small square. Penalise ‘blobs’ ≥ half a small square. Ecf allowed from table.

U2 Error bars in lg (T/s) plotted correctly.

All error bars must be plotted. Check first and last point. Must be accurate within half a small

square; penalise ≥ half a small square.

(ii) G2 Line of best fit If points are plotted correctly then lower end of line should pass between (1.65, 0.124) and (1.65, 0.128) and upper end of line should pass between (2.00, 0.300) and (2.00, 0.306). Allow ecf from points plotted incorrectly; five trend plots needed – examiner judgement.



(iii) C1 Gradient of best fit line


Work to half a small square; penalise ≥ half a small square.

U3 Uncertainty in gradient Method of determining absolute uncertainty Difference in worst gradient and gradient.

(iv) C2 y-intercept Must be negative. Check substitution of point from line into c = y – mx. Allow ecf from (c)(iii).



© UCLES 2010

U4 Uncertainty in y-intercept Method of determining absolute uncertainty Difference in worst y-intercept and y-intercept. Do not allow ecf from false origin read-off (FOX). Allow ecf from (c)(iv).

(d) C3 a = 10y-intercept y-intercept must be used. Expect an answer of about 0.19. If FOX expect answer of about 1.3.

C4 b = gradient and in the range 0.495 to 0.520 and to 2 or 3 sf

Allow 0.50 to 0.52 to 2 sf Penalise 1 sf or ≥4 sf

U5 Absolute uncertainty in a and b Difference in a and worst a. Uncertainty in b should be the same as the uncertainty in the gradient.

[Total: 15]


1. Uncertainty = gradient of line of best fit – gradient of worst acceptable line

2. Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient) (c) (iv) [U4]

1. Uncertainty = y-intercept of line of best fit – y-intercept of worst acceptable line

2. Uncertainty = ½ (y-intercept of steepest worst line – y-intercept of shallowest worst line)

(d) [U5]

1. Uncertainty = 10 best y-intercept - 10 worst y-intercept





9702 PHYSICS







© UCLES 2010

1 Planning (15 marks) Defining the problem (3 marks) P1 c, d or A is the independent variable and R is the dependent variable or vary c, d or A and

measure R. [1] P2 If c varied then (t and) d or A kept constant, if d varied then (t and) c or A kept constant, if A

varied then c or d kept constant. [1] P3 Keep temperature constant. [1] Methods of data collection (5 marks) M1 Circuit diagram to measure resistance. [1] M2 Use micrometer screw gauge to measure d or t. (Allow digital or vernier callipers) [1] M3 Measure c with a ruler/metre rule. [1] M4 Method of making contact with the strip e.g. use electrodes of at least same dimension as c

or d or t or conducting paint methods. Do not allow crocodile clips, unless it is clear that the whole area of the end of the strip is covered. [1]

M5 Method to determine resistance. [1] Method of analysis (2 marks) A1 Plot a graph of R against c, 1/d or 1/A depending on orientation. Other alternatives possible,

e.g. R against 1/c depending on orientation [1] A2 Must be consistent with A1: ρ = A × gradient or t × gradient/c [1] Other alternatives possible, e.g. ρ = d × gradient/t Safety considerations (1 mark) S1 Reference sharp edges or cutting metals, e.g. wear gloves. [1] Additional detail (4 marks) D1/2/3/4 Relevant points might include [4]

1. Insulate aluminium strip 2. Take many readings of t or d and average 3. Use a protective resistor/circuit designed to reduce current 4. Rearrange equation to determine graph using c, d and t or A 5. Determine typical resistance of aluminium strip 6. Likely meter range of ammeter/voltmeter/ohmmeter 7. Detail on cutting strip e.g. mark using set square


[Total: 15]



© UCLES 2010



(a) A1

C

t− Must be negative. Allow

C

15− .

(b) T1 T2

150 1.28 or 1.281

100 1.61 or 1.609

66.7 1.86 or 1.856

50.0 1.97 or 1.974

33.3 2.08 or 2.079

T1 for 1/R column – ignore sf and rounding errors T2 for ln (V/V) column – must be values given A mixture is allowed

U1 From ± 0.05 or ± 0.06 to ± 0.02 or

± 0.03


(c) (i) G1 Five points plotted correctly Must be within half a small square; penalise ≥ half a small square. Ecf allowed from table.

Penalise ‘blobs’ ≥ half a small square.

U2 Error bars in ln(V/V) plotted correctly.

All plots to have error bars; penalise ≥ half a small square. Check first and last point. Must be accurate within half a small square.

(ii) G2 Line of best fit If points are plotted correctly then upper end of line should pass between (20, 2.16) and (20, 2.18) and lower end of line should pass between (160, 1.20) and (160, 1.225). Allow ecf from points plotted incorrectly – examiner judgement.



(iii) C1 Gradient of best fit line Must be negative


Work to half a small square; penalise ≥ half a small square. Do not penalise POT.

U3 Uncertainty in gradient Method of determining absolute uncertainty. Difference in worst gradient and gradient.

(d) (i) C2 C = –15/gradient Gradient must be used. Allow ecf from (c)(iii). Do not penalise POT.

C3 2.14 × 10–3 F to 2.24 × 10–3 F and to 2 or 3 sf

Must be in range – penalise POT.

Allow equivalent unit including s Ω–1, C V–1, A s V–1



© UCLES 2010

(ii) U4 Determines % uncertainty in C Uses worst gradient or worst calculated C value. Do not check calculation.

(e) C4 Determines R correctly Expect to see an answer about 3000 Ω. R = 6.514/candidate’s C; allow ecf from (d)(i)

U5 Determines absolute uncertainty Determines worst value of R or (d)(ii) × R

[Total: 15]


1. Uncertainty = gradient of line of best fit – gradient of worst acceptable line 2. Uncertainty = ½ (steepest worst line gradient – shallowest worst line gradient)

(d) (ii) [U4]

1. Works out worst C then determines % uncertainty 2. Works out percentage uncertainty in gradient

(e) [U5]

1. Works out worst R then determines difference

2. RC

CRR

∆=

∆=∆

gradient

gradient

Education

9702 w10 ms_all